Conservation of Momentum - Shuffleboard

[Prixfex]

Hi PhysicsForums. :P First time here. I would just like someone to verify whether I did this problem correctly.

A moving shuffleboard puck has a glancing collision with a stationary puck of the same mass. If friction is negligible, what are the speeds of the pucks after the collision?

Given variables: v1i = 0.95 m/s, after collision, m1 (puck 1) moves at a 50 degree angle to the horizontal, positive direction in the y (assuming a regular coordinate grid), while m2 (puck 2) moves at a 40 degree angle toe the horizontal, negative direction in the y.

Attempt:

Conservation of Momentum (X direction)

Pi = Pf
m1v1ix + m2v2ix = m1v1fx + m2v2fx
v1ix = v2fx-v1fx
0.95 = v2 (cos 40) - v1 (cos 50)

Conservation of Momentum (Y direction)

m1v1iy + m2v2iy = m1v1fy + m2v2fy
0 = v1fy + v2fy
-v1 (sin 50) = v2 (sin 40)
v1 is a vector, so since its moving in a positive direction (according to my grid), it's still positive anyway...
v2 = v1(sin50)/sin40

substitute this into the equation I got for C.o.M. in the X

0.95 m/s = v1 (cos50) + v1(sin50)(cos40)/sin(40)
0.95 m/s = v1 (cos 50 + sin50(cos 40)/sin(40)

v1 = 0.95m/s / (cos 50 + sin50cos40/sin40)
v1 = .6106482 m/s

substitute this into the equation I had for C.o.M. in the Y

v2 = .6106482 m/s (sin 50)/sin40
v2 = .7277422 m/s

Hopefully I got this right? >_>

 

[lippyka]

Yes, you got it correct! The final speeds of the two pucks after the collision are 0.6106482 m/s and 0.7277422 m/s.

 

[kerimek]

Hello, welcome to PhysicsForums! Your approach to solving this problem is correct. You have correctly applied the conservation of momentum in both the x and y directions. Your final calculations for the speeds of the pucks are also correct. Great job!